Inequalities Review

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Inequalities Review

by amitdgr » Mon Aug 18, 2008 1:43 am
Points to remember :
1. if x > y then (1/x) < (1/y)
2. if x > y then -x < -y (that is the inequality gets reversed when both sides are multiplied by negative sign)
3. NEVER EVER , cross-multiply a variable (or) expression in an inequality blindly. You CAN cross multiply if and only if you are sure that the variable (or) expression being cross-multiplied is positive.
4. You can blindly cross multiply constant terms or numbers.

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For any real numbers, a; b,and c:
a < b is equivalent to a + c < b + c;
a > b is equivalent to a + c > b + c;
a = b is equivalent to a + c = b + c;
a >= b is equivalent to a + c >= b + c.
a <= b is equivalent to a + c <= b + c.

In other words, when we add or subtract the same number on both sides of an inequality, the direction of the inequality symbol is not changed.

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For any real numbers, a; b, and any positive number c:
a < b is equivalent to ac < bc;
a > b is equivalent to ac > bc.
a < b is equivalent to a/c < b/c;
a > b is equivalent to a/c > b/c.

For any real numbers, a; b, and any negative number c:
a < b is equivalent to ac > bc;
a > b is equivalent to ac < bc.
a < b is equivalent to a/c > b/c;
a > b is equivalent to a/c < b/c.

Similar statements hold for >= and <=

In other words, when we multiply or divide by a positive number on both sides of an inequality, the direction of the inequality symbol stays the same.

When we multiply or divide by a negative number on both sides of an inequality, the direction of the inequality symbol is reversed.

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|x| = x if x > 0
|x| = -x if x < 0
if |x| > y , then either x > y or -x > y
if |x| < y , then either x < y or -x < y

let "r" be a positive real number and "a" be a fixed real number, then
|x-a| < r implies a-r < x < a+r in other words x lies somewhere in between a-r and a+r
|x-a| > r implies x < a-r or x > a+r in other words, x lies outside a+r and a-r

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1 )if x^2 > k^2 and x> 0 then this implies x >k ( only for positive integers)

2) If k > 1 , then

a) k^x > 1 when x>0
b) 0< k^x < 1 when x < 0

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For word problems :

x is at least 30 implies x>=30 ( that is x is minimum 30)
x is at most 30 implies x<=30 ( that is x is maximum 30)
x cannot exceed 45 implies x <=45 ( that is x is maximum 45)
x must exceed 34 implies x > 34
x is between 7 and 12 implies 7 < x < 12
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Please add if i missed something.
Last edited by amitdgr on Thu Sep 11, 2008 8:54 pm, edited 2 times in total.
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by jazzcat4u » Mon Aug 18, 2008 8:16 am
...for the most part, i think our notes match up, but thought id add some anyway...

x is fewer than y................................... x < y
x has at least “some amount” as y......... x > y
there is more x than y.......................... x > y
x is greater than y................................ x > y
x has at most “some amount” as y........ x < y
x is no less than y................................ x > y
x is no greater than y........................... x < y

When you are asked about |x| or x² ALWAYS remember to consider –x and +x

When comparing x and x² whatever the value of x, the value of x² must be by definition, be at least 0, in otherwords, x² will always be positive not negative.
Last edited by jazzcat4u on Mon Aug 18, 2008 10:18 am, edited 1 time in total.

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by amitdgr » Mon Aug 18, 2008 9:29 am
jazzcat4u wrote:...for the most part, i think our notes match up, but thought id add some anyway...

x is fewer than y................................... x < y
x has at least “some amount” as y......... x > y
there is more x than y.......................... x > y
x is greater than y................................ x > y
x has at most “some amount” as y........ x < y
x is no less than y................................ x > y
x is no greater than y........................... x < y

When you are asked about |x| or x² ALWAYS remember to consider –x and +x
Great !! Bring in more ...
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by bha » Mon Aug 18, 2008 10:04 am
amitdgr/jazzcat4u
Thanks for sharing...

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by bha » Mon Aug 18, 2008 10:19 am
amitgr,
I am confused with following two statments:
if |x| > y , then either x > y or -x > y
I believed that only +x > y..we can't deduce anything abount -x??
if |x| < y , then either x < y or -x < y
same here...

Can you please explain with some examples...

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by amitdgr » Tue Aug 19, 2008 12:06 am
bha wrote:amitgr,
I am confused with following two statments:
if |x| > y , then either x > y or -x > y
I believed that only +x > y..we can't deduce anything abount -x??
if |x| < y , then either x < y or -x < y
same here...

Can you please explain with some examples...
I am not very sure :lol: :twisted: but I will try to explain...

Let us take |x| &#8805; 5, this means that the absolute value of x must be greater than or equal to 5. That is the value of x can be 5, 6, 7, 8 ... etc. or -5 , -6 , -7... etc. for the inequality to hold true. The inequality will fail for all values from -4 to +4

In other words x can be greater than or equal to 5, or can be less than or equal to &#8722;5.

We write x &#8804; &#8722;5 or x &#8805; 5.
This can be re-written as -x &#8805; 5 ( multiply -1 on both sides of x &#8804; &#8722;5 ) or x &#8805; 5

so if |x| &#8805; y then x &#8804; &#8722;y or x &#8805; y or we can say -x &#8805; y or x &#8805; y


Hope it explains. Please correct me if i am wrong. I am also a learner here :)
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by maihuna » Wed Apr 15, 2009 10:35 am
this is an amazing post. thank you.